Dimensional Analysis: Limiting Reagent, Theoretical Yield, Percent Yield, Excess Reactant

In this example problem, we use dimensional analysis to find the limiting reagent and calculate the theoretical yield, percent yield, and excess reagent remaining. Our problem reads, "85.1 grams chlorine is combined with 119 grams fluorine according to the reaction: Cl2 gas plus 3 F2 gas produces 2 ClF3 gas. Part A: calculate the theoretical yield of chlorine trifluoride in grams. Part B: if 172 grams chlorine trifluoride is actually produced from the reaction, calculate the percent yield. Part C: calculate the mass in grams of excess reactant remaining." In this problem, we're given amounts of two different reactants so we need to determine which one runs out first; this is a limiting reagent problem. To calculate the theoretical yield, we need to determine how much product can be produced from each amount of reactant. Whichever produces less product is the limiting reagent, and the amount it can produce is the theoretical yield. We'll use dimensional analysis to convert 85.1 grams chlorine gas into grams of chlorine trifluoride. We can use the molar mass of chlorine to convert from grams of chlorine to moles of chlorine. Then we can use the coefficients in the balanced chemical equation to convert from moles of chlorine to moles of chlorine trifluoride. Then we can use the molar mass of chlorine trifluoride to convert from moles of chlorine trifluoride to grams of chlorine trifluoride. Completing the calculation, we should get 222 grams of chlorine trifluoride. Next, we have to convert 119 grams of fluorine gas to grams of chlorine trifluoride so we know which can produce less product. We can use the molar mass of fluorine to convert from grams of fluorine to moles of fluorine. Then we can use the coefficients in the balanced chemical equation to convert from moles of fluorine to moles of chlorine trifluoride. Then we can use the molar mass of chlorine trifluoride to convert from moles of chlorine trifluoride to grams of chlorine trifluoride. Completing the calculation, we should get 193 grams of chlorine trifluoride. Since F2 produced less product, it is the limiting reagent; the amount of product it produced, 193 g ClF3, is the theoretical yield. Next, we want to calculate percent yield. Our equation for percent yield is actual yield divided by theoretical yield times 100%. We're told when the reaction is performed 172 grams of ClF3 is actually produced; this is the actual yield. We calculated the theoretical yield in part A: 193 grams of chlorine trifluoride. Completing the calculation, we get 89.1% as our percent yield. To determine how much excess reacted remains, we first need to determine how much of the excess reactant is needed to react with our limiting reagent. This will tell us how much of our excess reactant is used in the reaction. We start the dimensional analysis with our limiting reagent, 119 grams F2. We need to convert this into grams of excess reagent Cl2. We use the molar mass of fluorine gas to convert from grams of fluorine to moles of fluorine. We use the coefficients in the balanced chemical equation to convert from moles of fluorine to moles of chlorine gas. We use the molar mass of chlorine to convert from moles of chlorine gas to grams of chlorine gas. Completing the calculation, we should get 74 grams of Cl2, chlorine gas. This means 74 grams of Cl2 is required to completely react with our limiting reagent F2. To determine how much Cl2 is left over, we take 85.1 grams of Cl2 and subtract 74 g Cl2 to get 11.1 grams of Cl2. We started with 85.1 grams of Cl2, we used 74 g in the reaction, so there should be 11.1 grams of Cl2 remaining.